Problem: $\dfrac{d}{dx}\left(\dfrac{2x^2+x-3}{2x+7}\right)=$
Answer: $\dfrac{2x^2+x-3}{2x+7}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d d x ( 2 x 2 + x − 3 2 x + 7 ) = ( 2 x + 7 ) d d x ( 2 x 2 + x − 3 ) − ( 2 x 2 + x − 3 ) d d x ( 2 x + 7 ) ( 2 x + 7 ) 2 The quotient rule = ( 2 x + 7 ) ( 4 x + 1 ) − ( 2 x 2 + x − 3 ) ( 2 ) ( 2 x + 7 ) 2 Differentiate ( 2 x 2 + x − 3 ) & ( 2 x + 7 ) = 8 x 2 + 2 x + 28 x + 7 − 4 x 2 − 2 x + 6 ( 2 x + 7 ) 2 Expand = 4 x 2 + 28 x + 13 ( 2 x + 7 ) 2 \begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{2x^2+x-3}{2x+7}\right) \\\\ &=\dfrac{(2x+7)\dfrac{d}{dx}(2x^2+x-3)-(2x^2+x-3)\dfrac{d}{dx}(2x+7)}{(2x+7)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(2x+7)(4x+1)-(2x^2+x-3)(2)}{(2x+7)^2} \gray{\text{Differentiate }(2x^2+x-3)\text{ & }(2x+7)} \\\\ &=\dfrac{8x^2+2x+28x+7-4x^2-2x+6}{(2x+7)^2} ~~\gray{\text{Expand}} \\\\ &=\dfrac{4x^2+28x+13}{(2x+7)^2} \end{aligned} In conclusion, $\dfrac{d}{dx}\left(\dfrac{2x^2+x-3}{2x+7}\right)=\dfrac{4x^2+28x+13}{(2x+7)^2}$, or any other equivalent form.